EN हिं

NCERT Solutions for Class 5th Maths Chapter 9 Sunitha's mother shares the candies; dividing Hundreds, Tens and Ones in turn — Division Using Place Value

Book page 128–129 Updated on2026-09-19

Q1.
Sunitha's mother has 62 candies to be distributed equally among 5 children. How many candies would each child get? 1. 62 ÷ 5 → Divide 62 into 5 equal parts.
Answer

Each child gets 12 candies, and 2 candies are left over.

Think of the 62 candies as 6 bundles of ten and 2 loose candies.

6 Tens and 2 Ones = 62 5 children take 1 ten each. 1 ten is left, so break it into 10 Ones. 10 + 2 = 12 Ones to share 12 ÷ 5 = 2 each, 2 left over
Share the tens first, then break what is left into ones.
  1. Share the Tens. 6 Tens ÷ 5 = 1 Ten each, and 1 Ten is left.
  2. Regroup. That leftover 1 Ten becomes 10 Ones. With the 2 Ones already there, we now have 12 Ones.
  3. Share the Ones. 12 Ones ÷ 5 = 2 Ones each, and 2 Ones are left.
  4. Read the answer. Each child gets 1 Ten and 2 Ones, that is 12 candies. The remainder is 2.
      T O
5) 62 ( 1  2
  −5
  ──
  12  (Ones)
 −10
  ──
   2  (remainder)
Check by a second route: N = D × Q + R → 5 × 12 + 2 = 60 + 2 = 62 ✓
Answer: 12 candies each, 2 left over.
Q2.
2. 75 ÷ 8 → Divide 75 into 8 equal parts. Can we divide this into 8 equal parts without breaking them? What can we do?
Answer

75 ÷ 8 = 9 with remainder 3.

Can we share the 7 Tens among 8 without breaking them? No. There are only 7 tens and 8 children, so not even one ten each. Every child gets 0 Tens.

  1. Regroup everything into Ones. 7 Tens = 70 Ones. Add the 5 Ones already there: 75 Ones.
  2. Share the 75 Ones among 8. 8 × 9 = 72, which fits. 8 × 10 = 80 is too many.
  3. So each part gets 9, and 75 − 72 = 3 are left.
      T O
8) 75 ( 0  9
 −72  (Ones)
  ──
   3  (remainder)
Why the 0 in the Tens place: It records that nobody got any tens. We do not write it in the final answer (we say 9, not 09) but writing it while working keeps every digit in its correct column.
Check by a second route: 8 × 9 + 3 = 72 + 3 = 75 ✓
Q3.
3. 324 ÷ 3 → Divide 324 into 3 equal parts. Why do we put a 0 here?
Answer

324 ÷ 3 = 108, with no remainder.

Break 324 into place values first: 3 Hundreds + 2 Tens + 4 Ones.

  1. Share the Hundreds. 3 Hundreds ÷ 3 = 1 Hundred each. Nothing left over.
  2. Share the Tens. 2 Tens shared among 3 — not possible without breaking them. So everyone gets 0 Tens.
  3. Regroup. The 2 Tens become 20 Ones. With the 4 Ones already there, we have 20 + 4 = 24 Ones.
  4. Share the Ones. 24 Ones ÷ 3 = 8 Ones each. Nothing left over.
  5. Read the quotient down the columns: 1 Hundred, 0 Tens, 8 Ones = 108.
       H T O
3) 324 ( 1  0  8
  −3  (Hundreds)
  ──
   2  (Tens)
  −0
  ──
  24  (Ones)
 −24
  ──
  00

Why do we put a 0 here?

Because there really are zero Tens in each part. The 0 is not decoration — it is a number. If you leave it out, the 1 and the 8 slide together and you write 18. But 18 is far too small: 3 × 18 = 54, not 324. The 0 holds the Tens place open so the 1 stays a Hundred.
Check by a second route: 3 × 108 = 324 ✓  Or split: 324 = 300 + 24, and 300 ÷ 3 = 100, 24 ÷ 3 = 8, so 100 + 8 = 108. Same answer ✓
Q4.
4. 136 ÷ 6 → Divide 136 into 6 equal parts.
Answer

136 ÷ 6 = 22 with remainder 4.

Break 136 into 1 Hundred + 3 Tens + 6 Ones.

  1. Try the Hundreds. 1 Hundred among 6 — not possible. Each part gets 0 Hundreds.
  2. Regroup. 1 Hundred becomes 10 Tens. With the 3 Tens already there, we have 13 Tens.
  3. Share the Tens. 13 Tens ÷ 6 = 2 Tens each (6 × 2 = 12), and 1 Ten is left.
  4. Regroup again. That 1 Ten becomes 10 Ones. With the 6 Ones already there, we have 16 Ones.
  5. Share the Ones. 16 Ones ÷ 6 = 2 Ones each (6 × 2 = 12), and 4 Ones are left.
  6. Read the quotient: 0 Hundreds, 2 Tens, 2 Ones = 22, remainder 4.
       H T O
6) 136 ( 0  2  2
 −12  (Tens)
  ──
  16  (Ones)
 −12
  ──
   4  (remainder)
Check by a second route: N = D × Q + R → 6 × 22 + 4 = 132 + 4 = 136 ✓
Q5.
Can you tell just by looking at the divisor and dividend, how many digits the quotient would have? Look at the problems above and find this out. Explain your thoughts.
Answer

Yes. Compare the divisor with the front of the dividend.

The test, in two steps:

  1. Look at the first digit of the dividend. If the divisor is smaller than or equal to it, the quotient has the same number of digits as the dividend.
  2. If the divisor is bigger than that first digit, the quotient has one digit fewer.
ProblemDivisor vs first digitDigits in the quotientQuotient
324 ÷ 33 is not bigger than 33 digits108
136 ÷ 66 is bigger than 12 digits22
62 ÷ 55 is not bigger than 62 digits12
75 ÷ 88 is bigger than 71 digit9

Explain your thoughts — sample answer: I look at the first digit of the dividend. In 136 ÷ 6, the 1 stands for 1 Hundred, and 1 Hundred cannot be shared among 6 parts. So there is no digit in the Hundreds place of the answer, and the quotient must be a 2-digit number. In 324 ÷ 3, the 3 Hundreds can be shared among 3, so the answer does have a Hundreds digit and is a 3-digit number.

For a two-digit divisor, compare it with the first two digits. In 902 ÷ 16, we ask whether 16 fits into 90 — it does, so the quotient has 2 digits (56). In 2874 ÷ 14, 14 fits into 28, so the quotient has 3 digits (205).
Was this helpful?